# If A, B, C, D Are in Proportion , Then Prove that a 2 + a B + B 2 a 2 − a B + B 2 = C 2 + C D + D 2 C 2 − C D + D 2 - Algebra

Sum

If a, b, c, d   are in proportion , then prove that

[a^2 + ab + b^2]/[a^2 - ab + b^2] = [c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]

#### Solution

It is given that a, b, c, d are in proportion.

therefore a/b = c/d = k

⇒ a = bk , c = dk

[a^2 + ab + b^2]/[a^2 - ab + b^2] = [(bk)^2 + bk xx b + b^2]/[(bk)^2 - bk xx b + b^2 ] = [b^2(k^2 + k + 1)]/[b^2(k^2 + k + 1 )] = [ k^2+k+1 ]/[ k^2 - k +1 ]     ....(1)

[c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ] = [(dk)^2 + dk xx d + d^2]/[(dk)^2 - dk xx d + d^2 ] = [d^2(k^2 + k + 1)]/[d^2(k^2 - k + 1 )] = [ k^2+k+1 ]/[ k^2 - k +1 ]      ...(2)

From (1) and (2), we get

[a^2 + ab + b^2]/[a^2 - ab + b^2] = [c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]`

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#### APPEARS IN

Balbharati Mathematics 1 Algebra 9th Standard Maharashtra State Board
Chapter 4 Ratio and Proportion
Problem Set 4 | Q 9.3 | Page 78